help with parse error please<

[liunx]

Hi all.

I've always get parse error in folowing code:



$stars2="slike/strana/2star.gif";


if (isset($kategorija))
{
if($kategorija=="3 *" echo "kategorija: " . '<img src=http://www.htmlforums.com/archive/index.php/"'. $stars2 .'" border="0"><br><br></a>'."<br>\n";

}


What I wanna do is to check if variable "$kategorija" is set. If true, check if variable is "3 *", and if true echo something, and display image.

Please helpyou have quotes in there that are not needed. and you forgot a closing )


$stars2="slike/strana/2star.gif";


if (isset($kategorija)) {
if($kategorija=="3 *") {
echo "kategorija: <img src=http://www.htmlforums.com/archive/index.php/\"$stars2\" border=\"0\"><br><br></a><br>\n";
}
}